Question

在我的任务中，我们有两个问题：我们有一个2路组关联缓存。缓存总共有四个集合。主存储器由4K块组成，每块8个字，并使用字寻址。部分a）要求演示地址格式，我已经解决了word = 3 bit set = 2 bit和field = 7 bit。我遇到的问题是部分b）：计算从位置8到位置51循环3次的程序的命中率。换句话说，将其视为汇编语言程序，从位置8处的操作码运行到位置51m处的操作码然后循环回位置8它总共进行了三次这样的迭代。

现在根据我的理解，我已经完成了一个标准，通常会给出某种速度或命中率？我想知道我如何计算命中率，如果我不知道未命中率，未命中罚款缓存速度或任何东西？

Answer 1

我认为我们在同一个班级大声笑我在今晚到期时有完全相同的问题。无论如何，我做了一些研究，并在chegg上找到了类似问题的答案：

a. Given that memory contains 2K blocks of eight words.
2K can be distributed as 2K * 23 = 211* 23 = 214 so we have 14-bit addresses with 9 bits
in the tag field, 2 bits in the set field and 3 in the word field

b. First iteration of the loop:
    → Address 8 is a miss, then entire block brought into Set 1.9-15 are then hits.
    → 16 is a miss, entire block brought into Set 2, 17-23 are hits.
    → 24 is a miss, entire block brought into Set 3, 25-31 are hits.
    → 32 is a miss, entire block brought into Set 0, 33-39 are then hits.
    → 40 is a miss, entire block brought into Set 1 41-47 are hits.
    → 48 is a miss, entire block brought into Set 2, 49-51 are hits.

    For the first iteration of the loop, we have 6 misses, and 5*7 + 3 hits, or 38 hits.
    On the remaining iterations, we have 5*8+4 hits, or 44 hits each,for 88 more hits.
    Therefore, we have 6 misses and 126 hits, for a hit ratio of 126/132, or 95.45%.

希望这有帮助，祝你好运！

如何从2路组关联缓存计算缓存命中率

1 个答案: